Exam 40: One-Dimensional Quantum Mechanics

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If an atom in a crystal is acted upon by a restoring force that is directly proportional to the distance of the atom from its equilibrium position in the crystal,then it is impossible for the atom to have zero kinetic energy.

(True/False)

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Question 21

An electron with kinetic energy 2.80 eV encounters a potential barrier of height 4.70 eV.If the barrier width is 0.40 nm,what is the probability that the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J,m_el = 9.11 × 10^-31 kg,h = 1.055 × 10^-34 J ∙ s, h = 6.626 × 10^-34 J ∙ s)

(Multiple Choice)

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Question 22

Calculate the ground state energy of a harmonic oscillator with a classical frequency of 3.68 × 10¹⁵ Hz.(1 eV = 1.60 × 10^-19 J,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s)

(Multiple Choice)

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Question 23

A 10.0-g bouncy ball is confined in a 8.3-cm-long box (an infinite well).If we model the ball as a point particle,what is the minimum kinetic energy of the ball? (h = 6.626 × 10^-34 J ∙ s)

(Multiple Choice)

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Question 24

The energy of a proton is 1.0 MeV below the top of a 6.8-fm-wide energy barrier.What is the probability that the proton will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s)

(Multiple Choice)

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Question 25

An electron is confined in a harmonic oscillator potential well.A photon is emitted when the electron undergoes a 3→1 quantum jump.What is the wavelength of the emission if the net force on the electron behaves as though it has a spring constant of 9.6 N/m? (m_el = 9.11 × 10^-31 kg, c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19 J,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s)

(Multiple Choice)

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Question 26

An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick.What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s)

(Multiple Choice)

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Question 27

One fairly crude method of determining the size of a molecule is to treat the molecule as an infinite square well (a box)with an electron trapped inside,and to measure the wavelengths of emitted photons.If the photon emitted during the n = 2 to n = 1 transition has wavelength 1940 nm,what is the width of the molecule? (c = 3.00 × 10⁸ m/s,h = 6.626 × 10^-34J ∙ s, M_el = 9.11 × 10^-31 kg)

(Multiple Choice)

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Question 28

An electron is in an infinite square well (a box)that is 8.9 nm wide.What is the ground state energy of the electron? (h = 6.626 × 10^-34 J ∙ s,m_el = 9.11 × 10^-31 kg,1 eV = 1.60 × 10^-1⁹)

(Multiple Choice)

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Question 29

The energy of a particle in the second EXCITED state of a harmonic oscillator potential is 5.45 eV.What is the classical angular frequency of oscillation of this particle? (1 eV = 1.60 × 10^-19 J,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s)

(Multiple Choice)

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Question 30

An electron is confined in a harmonic oscillator potential well.What is the longest wavelength of light that the electron can absorb if the net force on the electron behaves as though it has a spring constant of 74 N/m? (m_el = 9.11 × 10^-31 kg,c = 3.00 × 10⁸ m/s, 1 eV = 1.60 × 10^-19 J,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s)

(Multiple Choice)

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Question 31

You want to have an electron in an energy level where its speed is no more than 66 m/s.What is the length of the smallest box (an infinite well)in which you can do this? (h = 6.626 × 10^-34 J ∙ s,m_el = 9.11 × 10^-31 kg)

(Multiple Choice)

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Question 32

Showing 21 - 32 of 32

Exam 40: One-Dimensional Quantum Mechanics

If an atom in a crystal is acted upon by a restoring force that is directly proportional to the distance of the atom from its equilibrium position in the crystal,then it is impossible for the atom to have zero kinetic energy.

An electron with kinetic energy 2.80 eV encounters a potential barrier of height 4.70 eV.If the barrier width is 0.40 nm,what is the probability that the electron will tunnel through the barrier? (1 eV = 1.60 × 10-19 J,mel = 9.11 × 10-31 kg,h = 1.055 × 10-34 J ∙ s, h = 6.626 × 10-34 J ∙ s)

Calculate the ground state energy of a harmonic oscillator with a classical frequency of 3.68 × 1015 Hz.(1 eV = 1.60 × 10-19 J,h = 1.055 × 10-34 J ∙ s,h = 6.626 × 10-34 J ∙ s)

A 10.0-g bouncy ball is confined in a 8.3-cm-long box (an infinite well).If we model the ball as a point particle,what is the minimum kinetic energy of the ball? (h = 6.626 × 10-34 J ∙ s)

The energy of a proton is 1.0 MeV below the top of a 6.8-fm-wide energy barrier.What is the probability that the proton will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, mproton = 1.67 × 10-27 kg,h = 1.055 × 10-34 J ∙ s,h = 6.626 × 10-34 J ∙ s)

An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick.What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10-19 J, mproton = 1.67 × 10-27 kg,h = 1.055 × 10-34 J ∙ s,h = 6.626 × 10-34 J ∙ s)

An electron is in an infinite square well (a box)that is 8.9 nm wide.What is the ground state energy of the electron? (h = 6.626 × 10-34 J ∙ s,mel = 9.11 × 10-31 kg,1 eV = 1.60 × 10-19)

The energy of a particle in the second EXCITED state of a harmonic oscillator potential is 5.45 eV.What is the classical angular frequency of oscillation of this particle? (1 eV = 1.60 × 10-19 J,h = 1.055 × 10-34 J ∙ s,h = 6.626 × 10-34 J ∙ s)

You want to have an electron in an energy level where its speed is no more than 66 m/s.What is the length of the smallest box (an infinite well)in which you can do this? (h = 6.626 × 10-34 J ∙ s,mel = 9.11 × 10-31 kg)

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Calculate the ground state energy of a harmonic oscillator with a classical frequency of 3.68 × 10¹⁵ Hz.(1 eV = 1.60 × 10^-19 J,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s)

A 10.0-g bouncy ball is confined in a 8.3-cm-long box (an infinite well).If we model the ball as a point particle,what is the minimum kinetic energy of the ball? (h = 6.626 × 10^-34 J ∙ s)

The energy of a proton is 1.0 MeV below the top of a 6.8-fm-wide energy barrier.What is the probability that the proton will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s)

An 80-eV electron impinges upon a potential barrier 100 eV high and 0.20 nm thick.What is the probability the electron will tunnel through the barrier? (1 eV = 1.60 × 10^-19 J, m_proton = 1.67 × 10^-27 kg,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s)

An electron is in an infinite square well (a box)that is 8.9 nm wide.What is the ground state energy of the electron? (h = 6.626 × 10^-34 J ∙ s,m_el = 9.11 × 10^-31 kg,1 eV = 1.60 × 10^-1⁹)

The energy of a particle in the second EXCITED state of a harmonic oscillator potential is 5.45 eV.What is the classical angular frequency of oscillation of this particle? (1 eV = 1.60 × 10^-19 J,h = 1.055 × 10^-34 J ∙ s,h = 6.626 × 10^-34 J ∙ s)

You want to have an electron in an energy level where its speed is no more than 66 m/s.What is the length of the smallest box (an infinite well)in which you can do this? (h = 6.626 × 10^-34 J ∙ s,m_el = 9.11 × 10^-31 kg)